lab 2 solution

8/9/2019 Lab 2 Solution

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CEE 377 Lab 2: Structural Systems Solution Winter 2015

This worksheet contains a mix of component and system modeling relevant to structural design.

1 Axial System Capacity

The figure below corresponds to a rearrangement of one of the simple systems you analyzed in HW 1. Themember forces (demands) have been determined and are reported in terms of the applied load, P . Thegeometric and material properties are as follows: E a = E b = E c = 29.000ksi; Aa = 0.5 in2, Ab = 0.75in2,and Ac = 1.0 in2; La = Lc = 24in, Lb = 18in.

P

a b c

P

0.2P

0.4P

0.4P

1. Determine the load, P y, sufficient to cause initial yielding in the structure. The yield stress is F y = 50 ksifor all the members. Member a:

f ay = AaF y = 0.5 in2 × 50ksi = 25 k

f a = 0.2P ⇒ 0.2P y = 25 k ⇒ P y = 125 k

Member b:f by = AbF y = 0.75in2 × 50 ksi = 37.5 k

f b = 0.4P ⇒ 0.4P y = 37.5 k ⇒ P y = 93.75 k

Member c:

f cy = AcF y = 1.0 in2 × 50ksi = 50 k

f c = 0.4P ⇒ 0.4P y = 50k ⇒ P y = 125 k

Member b governs: P y = 93.75 k ←. Note the load sharing analysis is only valid up to this load levelbecause it was based on an assumption of linear behavior.

2. Add arrowheads and value labels to the free body diagram below assuming all members have reachedfull yield.

AaF y AbF y AcF y

AaF y = 0.5 in2 × 50ksi = 25k

AbF y = 0.75in2 × 50 ksi = 37.5 k

AcF y = 1.0 in2 × 50ksi = 50k

P cap

3. Use the result above to obtain the failure capacity load, P cap, for this structure.

P cap = f ay + f by + f cy = 25k + 37.5 k + 50 k = 112.5 k

1



CEE 377 2

4. What is the ratio of the ultimate capacity, P cap, to the first yield capacity, P y?

P capP y

= 112.5 k

93.75 k = 1.2

5. What would the ratio of the ultimate capacity, P cap, to the first yield capacity, P y be in the case of asingle-member structure (i.e., just one rod)?

P y = P cap ⇒ P yP cap

= 1.0

Thus in this case the 3-member structure has a reserve capacity of 20% beyond first yield, while asingle member structure would have no reserve capacity. Note that factors of safety applied in eachcase would not represent consistent factors of safety against collapse.



CEE 377 3

2 MDOF Spring Systems

Consider the simple spring structure as shown below:

ka

kb

kc

k

d

P 1,∆1 P 2,∆2

We know the loads and displacements can be related through an expression of the form

{P } = [K ] {∆}

Because this is a 2-DOF system (two joints can move), we know further we can write

P 1P 2

=

K 11 K 12K 21 K 22

∆1

∆2

1. Use the direct physical stiffness approach to construct the system stiffness for this system column bycolumn.

(a) Apply a unit displacement at joint 1 and determine the corresponding joint loads (this gives thefirst column of the stiffness, to be filled in below).

(b) Repeat for the case of a unit displacement at joint 2.

Solution:

[K ] =

ka + kb + kc −kc

−kc kc + kd

2. Use the following numerical values to determine the corresponding numerical system stiffness: ka =100k/in, kb = 150 k/in, kc = 200 k/in, kd = 50 k/in.

Solution:

[K ] =

450 −200−200 250

3. Construct the system load matrix for the case of a 20 kip load applied at joint 1:

{P } =

P 1P 2

=

200

k



CEE 377 5

3 Beam Integration and Stiffness

We will be considering beams in isolation in this exercise, but it’s always good to remember that beams areusually parts of larger structural systems. The figure below shows on common context that could lead tosomething similar to the configuration to be analyzed here:

Lateral Loads (wind, earthquake, hblasti)

Tributary Areas

Gravity Loads (live loads, dead loads)

The boundary conditions for the beam in the illustration differ from what we will consider here, but avariety of boundary conditions are possible depending on framing decisions, the role of the beam in the

overall structural system, and other factors.

Beam Governing Equations

The governing equation for beams relates the displaced shape, v(x), to the applied load, w(x). For the signconvention in which v(x) is positive downwards, the beam equation can be stated as

EIv(x) = w(x)

Similarly, the other primary physical quantities of interest (moment, shear, and rotation) can be related tothe displaced shape through the following relations:

−EIv(x) = V (x)

−EIv(x) = M (x)

v(x) = θ(x)

Note for v(x) positive up (a common sign convention), the same expressions become:

EIv(x) = −w(x)

EIv(x) = V (x)

EIv(x) = M (x)

v(x) = θ(x)

Either sign convention can be used—the beam itself doesn’t know and won’t care.



CEE 377 6

Application Practice

1. Use the governing equation and direct integration to obtain an expression for the displacement of thebeam shown following the steps outlined below.

w0

EI,L

(a) Use indefinite integration to obtain an expression for v(x) in terms of EI , L, the applied load,w0, and four constants of integration.

Integration:

EIv = w0

EIv = w0x + c1

EIv = w0x2/2 + c1x + c2

EIv = w0x3/6 +

c12 x2 + c2x + c3

EIv = w0x4/24 + c1

6 x3 + c2

2 x2 + c3x + c4

(b) State the boundary conditions for this case

v(0) = 0

v(0) = 0

v(L) = 0

v(L) = 0

(c) Use the boundary conditions to determine the unknown constants from part (a). (Starting withthe conditions at x = 0 makes for the easiest algebra).

v(0) = 0 ⇒ c4 = 0

v(0) = 0 ⇒ c3 = 0

v(L) = 0 ⇒ 0 = w0L4

24 +

c1L3

6 +

c2L2

2

v

(L) = 0 ⇒ 0 = w0L

3

6 + c1L

2

2 + c2L⇒ c1L = −2c2 − w0L

2

3

Substitute c1-c2 relation into v(L) condition:

w0L4

24 −

2c2L2

6 −

w0L4

36 +

c2L2

2 = 0 ⇒ c2 =

w0L2

12 ⇒ c1 = −

w0L

2

Displacement expression:

v(x) = w0

EI

x4

24 − x3L

12 +

x2L2

24

= w0x

2

24EI (L− x)

2



CEE 377 7

(d) Check your displacement expression by evaluating the shear at x = 0, which in this case youshould be able to get from inspection using symmetry. Recall V (0) = −EIv(0) as given above,and note also you already have an expression for v (x) from part (a).

V (0) = −EIv(0) = −c1 = w0L

2

(e) Determine the moments at x = 0 and x = L/2, and use these values to sketch a labeled momentdiagram.

Left end:

M (0) = −EIv(0) = −c2 = −w0L

2

12

Use symmetry tog et right end:M (L) = M (0)

Center point

M (L/2) = −EIv(L/2) = −w0L2

8 −

c1L

2 − c2 = −

w0L2

8 +

w0L2

4 −

w0L2

12 =

w0L2

24

w0

EI , L

w0L2

24

−

w0L12

w0L

2



CEE 377 8

2. Determine an expression for the displaced shape of the beam shown, v(x), and determine the endmoments and end shears. There is no rotation at either end of the beam ( v(0) = v (L) = 0).

v0

EI,L

(a) Use indefinite integration to obtain an expression for v(x) in terms of EI , L, and four constantsof integration. Note w(x) = 0 in this case.

Integration (Because w(x) = 0, could drop the EI factor to simplify algebra)

EIv = 0

EIv = c1

EIv = c1x + c2

EIv = c1

2 x2 + c2x + c3

EIv = c1

6

x3 + c2

2

x2 + c3x + c4

(b) State the boundary conditions for this case For sign convention with positive v downwards:

v(0) = −v0

v(0) = 0

v(L) = 0

v(L) = 0

(c) Use the boundary conditions to determine the unknown constants from part (a).

v(0) = v0 ⇒ c4 = −EIv0

v(0) = 0 ⇒ c3 = 0

v(L) = 0 ⇒ 0 = c1L

3

6 +

c2L2

2 −EIv0

v(L) = 0 ⇒ 0 = c1L

2

2 + c2L⇒ c1L = −2c2

Substitute c1-c2 relation into v(L) condition:

−2c2L2

6 +

c2L2

2 −EIv0 = 0 ⇒ c2 =

6EI

L2 v0 ⇒ c1 = −

12EI

L3 v0

Displacement expression (note there is no EI ):

v(x) = −v0

2x3

L3 −

3x2

L2 + 1



CEE 377 10

4 Qualitative Moment Diagrams

In this section we will practice identification and qualitative construction of moment diagrams for simplebeams and frames. Remember the basic toolkit: sketch the displaced shape to relate curvature to moment;identify locations where the moment must vanish; look for symmetries; use the string analogy to determinelocally the shape of the diagram (e.e, curved/straight, concavity direction, etc.).

1. Select the correct moment diagram for the loaded beam shown from the choices below:

None of the above

2. Which moment diagram matches the frame loaded as shown?

None of these

lab 2 solution

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